Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as.

Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. Blaise Pascal later solved this problem.

Binomial Distribution
p = .482 of zero aces = .518 at least one ace will occur

Binomial Distribution
p = .508 of zero double aces = .492 at least one double ace will occur

Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as DeMere's problem, named after Chevalier De Mere. More likely at least one ace with 4 throws will occur

Example You give 100 random students a questionnaire designed to measure attitudes toward living in dormitories Scores range from 1 to 7 (1 = unfavorable; 4 = neutral; 7 = favorable) You wonder if the mean score of the population is different then 4

Hypothesis Alternative hypothesis H1: sample = 4
In other words, the population mean will be different than 4

Hypothesis Alternative hypothesis Null hypothesis H1: sample = 4
In other words, the population mean will not be different than 4

Results N = 100 X = 4.51 s = 1.94 Notice, your sample mean is consistent with H1, but you must determine if this difference is simply due to chance

Results N = 100 X = 4.51 s = 1.94 To determine if this difference is due to chance you must calculate an observed t value

Observed t-value tobs = (X - ) / Sx

Observed t-value tobs = (X - ) / Sx
This will test if the null hypothesis H0:  sample = 4 is true The bigger the tobs the more likely that H1:  sample = 4 is true

Observed t-value tobs = (X - ) / Sx Sx = S / N

Observed t-value tobs = (X - ) / .194 .194 = 1.94/ 100

Observed t-value tobs = (4.51 – 4.0) / .194

Observed t-value 2.63 = (4.51 – 4.0) / .194

t distribution

t distribution tobs = 2.63

t distribution tobs = 2.63 Next, must determine if this t value happened due to chance or if represent a real difference in means. Usually, we want to be 95% certain.

t critical To find out how big the tobs must be to be significantly different than 0 you find a tcrit value. Calculate df = N - 1 Page 747 First Column are df Look at an alpha of .05 with two-tails

t distribution tobs = 2.63

t distribution tcrit = -1.98 tcrit = 1.98 tobs = 2.63

t distribution Reject H0:  sample = 4 tcrit = -1.98 tcrit = 1.98
tobs = 2.63 If tobs fall in critical area reject the null hypothesis Reject H0:  sample = 4

t distribution Do not reject H0:  sample = 4 tcrit = -1.98
tobs = 2.63 If tobs does not fall in critical area do not reject the null hypothesis Do not reject H0:  sample = 4

Decision Since tobs falls in the critical region we reject Ho and accept H1 It is statistically significant, students ratings of the dorms is different than 4. p < .05

Example You wonder if the average IQ score of students at Villanova significantly different (at alpha = .05)than the average IQ of the population (which is 100). You sample the students in this room. N = 54 X = 130 s = 18.4

The Steps Try to always follow these steps!

Step 1: Write out Hypotheses
Alternative hypothesis H1: sample = 100 Null hypothesis H0: sample = 100

Step 2: Calculate the Critical t
N = 54 df = 53  = .05 tcrit = 2.0

Step 3: Draw Critical Region
tcrit = -2.00 tcrit = 2.00

Step 4: Calculate t observed
tobs = (X - ) / Sx

tobs = (X - ) / Sx Sx = S / N

tobs = (X - ) / Sx 2.5 = 18.4 / 54

tobs = (X - ) / Sx 12 = ( ) / 2.5 2.5 = 18.4 / 54

Step 5: See if tobs falls in the critical region
tcrit = -2.00 tcrit = 2.00

tcrit = -2.00 tcrit = 2.00 tobs = 12

Step 6: Decision If tobs falls in the critical region:
Reject H0, and accept H1 If tobs does not fall in the critical region: Fail to reject H0

Step 7: Put answer into words
We reject H0 and accept H1. The average IQ of students at Villanova is statistically different ( = .05) than the average IQ of the population.

Practice You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average average paranoia score significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

Scores

Alternative hypothesis H1: sample = 56.1 Null hypothesis H0: sample = 56.1

N = 5 df =4  = .10 tcrit = 2.132

tcrit = tcrit = 2.132

tobs = (X - ) / Sx -.48 = ( ) / 1.88 1.88 = 4.21/ 5

tcrit = tcrit = 2.132 tobs = -.48

We fail to reject H0 The average paranoia of your friends is not statistically different ( = .10) than the average paranoia of the population.

One-tailed test In the examples given so far we have only examined if a sample mean is different than some value What if we want to see if the sample mean is higher or lower than some value This is called a one-tailed test

Remember You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average paranoia score significantly ( = .10) different than the average paranoia of the population ( = 56.1)?

Hypotheses Alternative hypothesis Null hypothesis H1: sample = 56.1

What if. . . You recently finished giving 5 of your friends the MMPI paranoia measure. Is your friends average paranoia score significantly ( = .10) lower than the average paranoia of the population ( = 56.1)?

Hypotheses Alternative hypothesis Null hypothesis
H1: sample < 56.1 Null hypothesis H0: sample = or > 56.1

N = 5 df =4  = .10 Since this is a “one-tail” test use the one-tailed column Note: one-tail = directional test tcrit = If H1 is < then tcrit = negative If H1 is > then tcrit = positive

tcrit =

tobs = (X - ) / Sx

tobs = (X - ) / Sx -.48 = ( ) / 1.88 1.88 = 4.21/ 5

tcrit =

tcrit = tobs = -.48

We fail to reject H0 The average paranoia of your friends is not statistically less then ( = .10) the average paranoia of the population.

Practice You just created a “Smart Pill” and you gave it to 150 subjects. Below are the results you found. Did your “Smart Pill” significantly ( = .05) increase the average IQ scores over the average IQ of the population ( = 100)? X = 103 s = 14.4

Alternative hypothesis H1: sample > 100 Null hypothesis H0: sample < or = 100

N = 150 df = 149  = .05 tcrit = 1.645

tcrit = 1.645

tobs = (X - ) / Sx 2.54 = ( ) / 1.18 1.18=14.4 / 150

tcrit = 1.645 tobs = 2.54

We reject H0 and accept H1. The average IQ of the people who took your “Smart Pill” is statistically greater ( = .05) than the average IQ of the population.

So far. . . We have been doing hypothesis testing with a single sample
We find the mean of a sample and determine if it is statistically different than the mean of a population

Basic logic of research

Start with two equivalent groups of subjects

Treat them alike except for one thing

See if both groups are different at the end

Notice This means that we need to see if two samples are statistically different from each other We can use the same logic we learned earlier with single sample hypothesis testing

Example You just invented a “magic math pill” that will increase test scores. You give the pill to 4 subjects and another 4 subjects get no pill You then examine their final exam grades

Hypothesis Two-tailed
Alternative hypothesis H1: pill = nopill In other words, the means of the two groups will be significantly different Null hypothesis H0: pill = nopill In other words, the means of the two groups will not be significantly different

Hypothesis One-tailed
Alternative hypothesis H1: pill > nopill In other words, the pill group will score higher than the no pill group Null hypothesis H0: pill < or = nopill In other words, the pill group will be lower or equal to the no pill group

For current example, lets just see if there is a difference
Alternative hypothesis H1: pill = nopill In other words, the means of the two groups will be significantly different Null hypothesis H0: pill = nopill In other words, the means of the two groups will not be significantly different

Results Pill Group 5 3 4 No Pill Group 1 2 4 3

Remember before. . . Step 2: Calculate the Critical t
df = N -1

Now Step 2: Calculate the Critical t
df = N1 + N2 - 2 df = = 6  = .05 t critical = 2.447

Remember before. . . Step 4: Calculate t observed
tobs = (X - ) / Sx

Now Step 4: Calculate t observed
tobs = (X1 - X2) / Sx1 - x2

tobs = (X1 - X2) / Sx1 - x2 X1 = 3.75 X2 = 2.50

tobs = (X1 - X2) / Sx1 - x2

Standard Error of a Difference
Sx1 - x2 When the N of both samples are equal If N1 = N2: Sx1 - x2 = Sx12 + Sx22

Results Pill Group 5 3 4 No Pill Group 1 2 4 3

Standard Deviation S = -1

Standard Deviation Pill Group 5 3 4 No Pill Group 1 2 4 3 X2= 10

Standard Deviation Pill Group 5 3 4 No Pill Group 1 2 4 3 X2= 10
Sx= .48 Sx= . 645

Sx1 - x2 When the N of both samples are equal If N1 = N2: Sx1 - x2 = Sx12 + Sx22

Sx1 - x2 When the N of both samples are equal If N1 = N2: Sx1 - x2 = (.48)2 + (.645)2

Sx1 - x2 When the N of both samples are equal If N1 = N2: Sx1 - x2 = (.48)2 + (.645)2 = .80

Standard Error of a Difference Raw Score Formula
When the N of both samples are equal If N1 = N2: Sx1 - x2 =

X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 Sx1 - x2 =

X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 Sx1 - x2 = 10 15

Sx1 - x2 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15

Sx1 - x2 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15
4 (4 - 1)

Sx1 - x2 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15
56.25 30 25 4 4 12

.80 = X1= 15 X12= 59 N1 = 4 X2= 10 X22= 30 N2 = 4 10 15 59
56.25 30 25 7.75 4 4 12

tobs = (X1 - X2) / Sx1 - x2 Sx1 - x2 = .80 X1 = 3.75 X2 = 2.50

Sx1 - x2 = .80 X1 = 3.75 X2 = 2.50

1.56 = ( ) / .80 Sx1 - x2 = .80 X1 = 3.75 X2 = 2.50

tcrit = tcrit = 2.447 tobs = 1.56

We fail to reject H0. The final exam grades of the “pill group” were not statistically different ( = .05) than the final exam grades of the “no pill” group.

What if The two samples have different sample sizes (n)

Results Psychology 110 150 140 135 Sociology 90 95 80 98

Results Psychology 110 150 140 135 Sociology 90 95 80

If samples have unequal n
All the steps are the same! Only difference is in calculating the Standard Error of a Difference

When the N of both samples is equal If N1 = N2: Sx1 - x2 =

When the N of both samples is not equal If N1 = N2: N1 + N2 - 2

Results Psychology 110 150 140 135 Sociology 90 95 80 X1= 535
N1 = 4 X2= 265 X22= 23525 N2 = 3

X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3 N1 + N2 - 2

X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3 265 535 N1 + N2 - 2

X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3 265 535 23525 72425 N1 + N2 - 2

X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3 265 535 23525 72425 3 4 4 3

X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3 265 535 23525 72425 3 4 4 3 5

X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3 265 535 23525 72425 (.58) 3 4 4 3 5

= 10.69 X1= 535 X12= 72425 N1 = 4 X2= 265 X22= 23525 N2 = 3
72425 114.31 3 4 4 3 5 = 10.69

Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as.

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Practice Which is more likely: at least one ace with 4 throws of a fair die or at least one double ace in 24 throws of two fair dice? This is known as.

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